Transcript of "Quantitative Techniques random variables"
1.
Mr. Rohan Bhatkar
2.
Content-
Random Variables
Probability Mass Function
Discrete Random Variables
Distribution Function
Variance
Expectation
Continuous Random Variables
3.
Random Variables
Consider an experiment of throwing two dice. We know that this experiments
has 36 outcomes .
Let x be the sum of numbers on the uppermost faces.
For ExampleThe value of x is 2 if outcome of the experiment is (1,1) & probability of this
outcome is 1/36.
In this case we can associate real number with each outcome of random
experiment or group of outcomes of random experiment.
Here X is called Random Variables.
4.
Probability Mass Function
If x is discrete random variables values x1, x2, x3, …., xn then probability of each value is described by a function called
the Probability Mass Function. The probability that random variable X takes values xi is denoted by p(xi).
IllustrationSuppose fair coin is marked 1 & 2, dice numbered 1, 2, 3, 4, 5 & 6 are thrown simultaneously then probability mass
function of random variables X which is sum of numbers on coin & dice is obtained as under.
The sample space is:
S={(1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (2,1), (2,2), (2,3), (2,4), (2,5), (2,6)}
Note that n(s) = 12
X: sum of numbers on coin & dice.
X = 2, 3, 4, 5, 6, 7, 8
P[x=2] = {(1,1)} = 1/12
P[x=3] = {(1,2)(2,1)} = 2/12
P[x=4] = {(1,3)(2,2)} = 2/12
P[x=5] = {(1,4)(2,3)} = 2/12
P[x=6] = {(1,5)(2,4)} = 2/12
P[x=7] = {(1,6)(2,5)} = 2/12
P[x=8] = {(2,6)} = 1/12
Thus probability distribution of random variable X is as under:
X=x
2
3
4
5
6
7
8
P[X=x]
1/12
2/12
2/12
2/12
2/12
2/12
1/12
Note that P[xi] ≥ 0 & ∑ (xi) = 1
5.
Discrete Random Variable
A random variable which can assume only a countable number of real values &
the values taken by variables depends on outcome of random experiment is
called Discrete Random Variables.
For Example1. Number of misprints per page of book.
2. Number of heads in n tosses of fair coin.
3. Number of throws of dice to get first 6.
6.
Distribution Function
Let X be a random variable. The function F defined for all real values x by F(x) = P [X = x], For all real x is called
Distribution Function.
A distribution function is also called as Cumulative Probability Distribution Function.
Properties OF Distribution Function:
If F is the D.F. of random variable X & if a <b then
P[a < X ≤ b] = F(b) – F(a).
Values of all distribution functions lie between 0 & 1
i.e. 0 ≤ F(x) ≤ 1 for all x.
All distribution functions are monotonically non-decreasing
i.e. 0 < y then F(x) < F(y).
F(- ∞) = lim F(x) = 0
x-∞
F(+ ∞) = lim F(x) = 1
x→∞
If X is discrete random variable then F(x) = ∑ P(xi)
xi ≤ x
If values of discrete random variable X are like x1 < x2 < x3 < x4 … then P(Xn+1) = F(Xn+1) – F(Xn)
If x is discrete random variable then D.E. is step function.
7.
IllustrationConsider probability distribution of random variable x
X=x
1
2
3
4
5
6
P[X=x]
0.1
0.2
0.3
0.2
0.1
0.1
If F(x) is distribution function of random variable x then
F(1) = P[x ≤ 1] = P(1) = 0.1
F(2) = P[x ≤ 1] = P(1) + P(2)
= 0.1 + 0.2 = 0.3
F(3) = 0.1 + 0.2 + 0.3 = 0.6
F(4) = 0.1 + 0.2 + 0.3 + 0.2 = 0.8
F(5) = 0.1 + 0.2 + 0.3 + 0.2 + 0.1 = 0.9
F(6) = 1
Thus values of x & corresponding cumulative probability distribution function is as under:
X=x
1
2
3
4
5
6
P[X=x]
0.1
0.3
0.6
0.8
0.9
1.0
We will have graphical representation of random variable X, & the Graph of D.E.
1.0
Probability Distribution
0.8
3
P(x)
f(x)
2
Cumulative
Probability Distribution
0.6
0.4
1
0.2
1
2
3
4
5
6
1
2
3
X→
X→
4
5
6
8.
Variance
If X is discrete random variable then variance of x is given by
Var (X) = E[X – X (X)]2
Note that Var (X) = µ2
Var (X) = µ2` - µ1`2
Var (X) = E (X2) – [E(X)] 2
The positive square root of the variance is known as standard deviation
S D (X) = + √Var (X)
IllustrationFor a random variable X, E (X) = 10 and Var (X) = 5. Find Var (3X + 5), Var(X – 2), Var (4X).
Also find E(5X – 4), E(4X + 3).
SolutionVar (3X + 5) = Var (3X)
= 9 Var (X) = 9 x 5 = 45
Var (x - 2) = Var (X) = 5
Var (4X) = 16 Var (X) = 16 x 5 = 80
Var (5X - 4) = 5 E(X) – 4
Var (5X - 4) = 5 x 10 – 4 = 46
Var (4X + 3) = 4 E(X) + 3
Var (4X + 3) = 4 x 10 + 3 = 43
9.
Properties OF Variance:
1. Variance is independent of change of origin. It means that if X is random variable & ‘a’ is constant then variances
of X and new variable X + a are same i.e.
Var (X) + Var (X + a)
consider
Var (X + a) = E [X + a – (E (X) + a)]2
Var (X + a) = E [X + a – E (X) + a]2
Var (X + a) = E [X –E (X)]2
Var (X + a) = Var (X).
2. Variance of random variable depends upon change of scale.
i.e. Var (a X) = a2 Var (X)
consider
Var (a X) = E [aX – E (aX)]2
= E [aX – a E (X)]2
= E a2 [X –E (X)]2
= a2 E [X –E (X)]2
= a2 Var (X).
3. From property 1 and 2 we can find Var (a X + b),
Var (a X + b) = Var (a X)
by property 1
2 Var (X)
=a
by property 2
4. Variance of constant is 0. put a=0 in a x + b then
Var (a X + b)
= Var (b)
but
Var (a x + b) = a2 × Var (X)
Var (a x + b) = 0 × Var (X) = 0
Var (b) = 0
10.
Expectation
Mathematical Expectation of Discrete Random Variable:
Once we have determined probability distribution function P(x) & distribution function of discrete
random variable X, we want to compute the mean or variance of random variable X, the mean or expected value of X
is nothing but weighted average of value X where corresponding probabilities are taken as weights, thus if X takes
values x1, x2,.. With corresponding probabilities p(x1), p(x2),… then mathematical expectation of X denoted by E(X) is
given by E(X) = ∑ xi P(xi).
E(X) is also mean of random variable X.
E(X) exists if series on right hand side is absolutely convergent.
IllustrationThe p.m.f. of a random variable X is given below. Find E(X).
Hence E(2X + 5) & E(X - 5).
X=x
1
2
3
4
5
6
P[X=x]
0.1
0.15
0.2
0.3
0.15
0.1
SolutionExpected value of random variable is given by:
E(X) = ∑xi P[X = xi]
= 1(0.1) + 2(0.15) + 3(0.2) + 4(0.3) + 5(0.15) + 6(0.1)
= 0.1 + 0.3 + 0.6 + 1.2 + 0.75 + 0.6
E(X) = 3.55
E(2X + 5) = 2 E(X) + 5
= 2 E(3.55) + 5
= 12.1
E(X - 5) = E(X) - 5
= (3.55) - 5
= -1.45
11.
Properties OF Expectation:
If X1 & X2 are two random variables then E(X1 + X2) = E(X1 ) + E(X2).
This result can be generalized for X1, X2, …, Xn i.e. n random variables
E(X1 + X2 + …. + Xn ) = E(X1) + E(X2) + …. + E(Xn)
If X & Y are independent random variable, E(XY) = E(X) E(Y)
If X is a random variable & a is constant then
E(a X) = a E(X) and
E(X + a) = E(X) + a
If X is a random variable and a & b are constants
E[a (X) + b] = a E(X) + b
If X is a random variable and a & b are constants & g (X) a function X is random variable then
E[a g (X) + b] = a E[g (X)] + b
If X1, X2, X3 ….,Xn are any n random variable and if a1, a2, ….,an are any n constants then
a1 X1 + a2 X2 + …. An Xn is called linear combination of n variables & expectation of linear combination is given by
E(a1 X1 + a2 X2 + ….. + an Xn)
=
=
=
If X ≥ 0 then E(X) ≥ 0.
If X and Y are two random variables & if X ≤ Y then E(X) ≥ E(Y).
If X & Y are independent random variable then
E[a g (X) + h (Y)] = E[g (X)] E[h (Y)]
then g (X) is a function of X & its random variable. Also h (Y) is a function of Y & random variable.
12.
Continuous Random Variables
Consider the small interval (X, X + dx) of length dx round the point x.
Let f(x) be any continuous function of x so that f(x) represent the probability that falls in very small
interval (X, x + dx)
Symbolically P[x ≤ X ≤ x + dx] = f(x) dx.
y
f(x) dx
Y = f(x)
x
X – dx/2 X + dx/2
In the figure f(x) dx represents the area bounded by the curve y = f(x); X-axis a& the ordinates x and x
+ dx.
The function f(x) so defined is known as probability density function of random variable X & usually
abbreviated as p.d.f. The expression f(x) dx usually written as F (x), is known as probability
differential & f(x) is known as probability density curve. The probability that X lie in the interval dx is
f(x) dx. Thus the p.d.f. of random variable x is defined as
f(x) = lim P [x ≤ X ≤ x + δx]
δx → 0
δx
13.
IllustrationA random variable X has following p.d.f.
f(x) = K
-∞<x<∞
2
1+x
=0
Otherwise
Find k:
Solution:
Since X is continuous random variable with density function f(x),
∞
⌠ f(x) dx
=1
-∞
∞
⌠
-∞
k[tan-1
K
dx
2
1+x
∞
x]
-∞
=1
=1
k[tan-1 ∞ - tan-1 ∞ (- ∞)] = 1
k[π/2 + (π/2)]
k
k
π
=1
=1
= 1/ π
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